Infinite square well expectation value problem

AI Thread Summary
The discussion revolves around calculating the expectation value of the operator 1/2<xp+px> for a particle in an infinite square well in the first excited state (n=2). Participants clarify the correct wave function, which is √(2/L) Sin((nπx)/L), and confirm the integration bounds from 0 to L. The initial attempts at setting up the expectation value formula are critiqued, with emphasis on the need to properly define the operators and the wave function. A suggestion is made to simplify the calculation by using the commutation relation [x,p] = ihbar, which aids in finding the expectation value. The conversation concludes with a reminder to verify the integral and calculations for accuracy.
Fakestreet123
Messages
2
Reaction score
0

Homework Statement


A particle in an infinite box is in the first excited state (n=2). Obtain the expectation value 1/2<xp+px>


2. The attempt at a solution

Honestly, I don't even know where to begin.
I assumed V<0, V>L is V=∞ and 0<V<L is V=0

I tried setting up the expectation value formula

1/2∫x|ψ*Pψ|2+(x|ψ2)*P x|ψ|2

but what is ψ? is it √(2/L) Sin ((n2πx)/L) because its the solution to the infinite well? are the bounds from 0 to L?

Thanks for viewing my question!
 
Last edited:
Physics news on Phys.org
\psi is the wave function for the given problem, that you find by solving Schrodinger equation for a given potential
 
Fakestreet123 said:

Homework Statement


A particle in an infinite box is in the first excited state (n=2). Obtain the expectation value 1/2<xp+px>


2. The attempt at a solution

Honestly, I don't even know where to begin.
I assumed V<0, V>L is V=∞ and 0<V<L is V=0
You mean V=∞ when x<0 or x>L.

I tried setting up the expectation value formula

1/2∫x|ψ*Pψ|2+(x|ψ2)*P x|ψ|2
How did you get this?

but what is ψ? is it √(2/L) Sin ((n2πx)/L) because its the solution to the infinite well? are the bounds from 0 to L?
You do use a solution for the infinite square well, but the function you wrote isn't quite correct.
 
Thanks for responding guys! I very much appreciate it!

vela said:
You mean V=∞ when x<0 or
x
>L.

Ahh yes, sorry for the crappy notation


How did you get this?

Uhhh Now that I've done some more studying I can see that its wrong but I was, at the time, hoping that the expectation values would be ∫<xp>+<px> = ∫<xp+px> and the expectation values of xp, px will just add together like magic.

Which is wrong! So I've redone the problem in an attempt to simply find the operator first

<xp> = -ihx(dψ/dx) and
<px> = -ih (d(xψ)/dx)
1/2<xp+px> = (-ih)/2 (x(dψ/dx)+(d(xψ)/dx)

Now the expectation value should be <1/2(xp+px)>= ∫ψ*(1/2)<xp+px>ψ dx?

do I just plug in 1/2<xp+px> (assuming its correct) and plug in ψ (the solution to the infinite square well) and take the integral from 0 to L?

You do use a solution for the infinite square well, but the function you wrote isn't quite correct.

Sorry! I was very wrong haha
√(2/L) Sin ((nπx)/L)
 
One thing that might help you:

Using [x,p] = ihbar, or xp - px = ihbar, or px = -ihbar + xp, you can write

xp + px = xp + xp - ihbar = 2xp - ihbar, which has an easier to calculate expectation value. (you should check my work, I may have made a mistake)

In general an expectation value is:

<Q> = ∫ψ*Qψdx, where Q on the right is the operator for the observable you want to calculate. ψ is the wavefunction of the particle, which should be supplied by your textbook (it looks like you have the right one though).

Removed by moderator[/color]

Where I used the normalization condition ∫ψ*ψdx = 1.

I'll leave it to you do actually do the integral, and to check my work.
 
Last edited by a moderator:
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top